1-1 phần 100
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(100-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}\right)\)
\(=(1-1)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}+\frac{2}{3}...+\frac{99}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Làm lại đề cho:
\(\left(\frac{1}{4}-1\right)\cdot\left(\frac{1}{9}-1\right)\cdot\left(\frac{1}{16}-1\right)...\left(\frac{1}{81}-1\right)\cdot\left(\frac{1}{100}-1\right)\)
Tính nhẩm
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1:1/10=10
1 gấp 10 lần 1/10
1/10:1/100=10
1 /10 gấp 10 lần 1/100
![](https://rs.olm.vn/images/avt/0.png?1311)
a
\(A=1+3+3^2+3^3+....+3^{100}\)
\(3A=3+3^2+3^3+3^4+.....+3^{101}\)
\(2A=3^{101}-1\)
\(A=\frac{3^{101}-1}{2}\)
b
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(B=1-\frac{1}{2^{99}}\)
c
\(C=5^{100}-5^{99}+5^{98}-5^{97}+....+5^2-5+1\)
\(5C=5^{101}-5^{100}+5^{99}-5^{98}+....+5^3-5^2+5\)
\(6C=5^{101}+1\)
\(C=\frac{5^{101}+1}{6}\)
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)
\(\Rightarrow\frac{1}{2}B=\)\(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{100}\)
\(\Rightarrow B-\frac{1}{2}B=\left[\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\right]-\left[\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{100}\right]\)
\(\Rightarrow\frac{1}{2}B=\frac{1}{2}-\left(\frac{1}{2}\right)^{100}\Rightarrow B=\left[\frac{1}{2}-\left(\frac{1}{2}\right)^{100}\right].2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
=99/100 nhé